MHB Why does the implication hold?

[evinda]

Hi! (Smile)

I am looking at the proof of the following sentence:

For each natural number $n$ it holds that $n \notin n$.

Proof :

We define the set $X=\{ n \in \omega: n \notin n\}$.
It suffices to show that $X$ is an inductive set, because then $X=\omega$.
Obviously $\varnothing \in X$.
We suppose that $n \in X$ and we will show that $n'=n \cup \{n\} \in X$.

We suppose that $n' \in n'$. Then $n \cup \{ n \} \in n \cup \{ n \}$.
We have two cases:

• $n \cup \{ n \} \in n$
• $n \cup \{ n \} \in \{ n \} \rightarrow n \cup \{ n \}=n$

If $n \cup \{ n \} \in n$ then $n \cup \{ n \} \subset n \rightarrow \{ n \} \subset n \rightarrow n \in n$, contradiction since $n \in X$.

If $n \cup \{ n \}=n \rightarrow n \cup \{ n \} \subset n$ and from the proof of the previous sentence we conclude again to a contradiction.

So, $n' \notin n'$ and so $X$ is inductive, i.e. $X=\omega$.Could you explain me why $n \cup \{ n \} \subset n \rightarrow \{ n \} \subset n \rightarrow n \in n$ ? (Thinking)

 

[Evgeny.Makarov]

Could you explain me why $n \cup \{ n \} \subset n \rightarrow \{ n \} \subset n \rightarrow n \in n$ ?

For all sets $A$, $B$ and $C$ and for all $x$ it is the case that $A\cup B\subseteq C\implies B\subseteq C$ and $\{x\}\subseteq A\iff x\in A$.

 

[evinda]

For all sets $A$, $B$ and $C$ and for all $x$ it is the case that $A\cup B\subseteq C\implies B\subseteq C$ and $\{x\}\subseteq A\iff x\in A$.

I understand... Thanks a lot! (Cool)